bmzctf 刷题 elrond32

直接看mian函数

跟进sub_8048414函数

signed int __cdecl sub_8048414(_BYTE *a1, int a2)
{
  signed int result; // eax

  switch ( a2 )
  {
    case 0:
      if ( *a1 == 105 )
        goto LABEL_19;
      result = 0;
      break;
    case 1:
      if ( *a1 == 101 )
        goto LABEL_19;
      result = 0;
      break;
    case 3:
      if ( *a1 == 110 )
        goto LABEL_19;
      result = 0;
      break;
    case 4:
      if ( *a1 == 100 )
        goto LABEL_19;
      result = 0;
      break;
    case 5:
      if ( *a1 == 97 )
        goto LABEL_19;
      result = 0;
      break;
    case 6:
      if ( *a1 == 103 )
        goto LABEL_19;
      result = 0;
      break;
    case 7:
      if ( *a1 == 115 )
        goto LABEL_19;
      result = 0;
      break;
    case 9:
      if ( *a1 == 114 )
LABEL_19:
        result = sub_8048414(a1 + 1, 7 * (a2 + 1) % 11);
      else
        result = 0;
      break;
    default:
      result = 1;
      break;
  }
  return result;
}

由前面的mian函数中的sub_8048414(a2[1], 0)可知，传入的a2为0，所以进入的是case 为0的环节，判断了一下指针a1后，进入了LABEL_19，在这里a2 = 7 * (a2 + 1) % 11，a2也就变成了7，然后再次进入sub_8048414，进入下一个switch轮回

这里可以写一段脚本算出来a2

a2 = 0
for i in range(9):    
    a2=7*(a2+1)%11
    print (a2)

得到a2为7,1,3,6,5,9,4,2,10

因为switch里的选择只到了9，所以10不要，所以range应该为8轮

相应的就可以得到a1值，转为char型为isengard

这里可以把上一步和这一步合并一下

s='i'
a2 = 0
key = 'ie ndags r'
for i in range(8):    
    a2=7*(a2+1)%11
    s += key[a2]
print (s)

可以得出程序需要的参数

运行一下

可以得到flag

根据伪代码求flag

但是也可以根据程序内的伪代码，算出flag

int __cdecl sub_8048538(int a1)
{
  int v2[33]; // [esp+18h] [ebp-A0h]
  int i; // [esp+9Ch] [ebp-1Ch]

  qmemcpy(v2, &dword_8048760, sizeof(v2));
  for ( i = 0; i <= 32; ++i )
    putchar(v2[i] ^ *(a1 + i % 8));
  return putchar(10);
}

其中dword_8048760地址处的值为

0x0F,0x1F,0x04,0x09,0x1C,0x12,0x42,0x09,0x0C,0x44,0x0D,0x07,0x09,0x06,0x2D,0x37,0x59,0x1E,0x00,0x59,0x0F,0x08,0x1C,0x23,0x36,0x07,0x55,0x02,0x0C,0x08,0x41,0x0A,0x14

exp:

a1 = [105,115,101,110,103,97,114,100]
flag=''
v2 =[0x0F,0x1F,0x04,0x09,0x1C,0x12,0x42,0x09,0x0C,0x44,0x0D,0x07,0x09,0x06,0x2D,0x37,0x59,0x1E,0x00,0x59,0x0F,0x08,0x1C,0x23,0x36,0x07,0x55,0x02,0x0C,0x08,0x41,0x0A,0x14]
for i in range(33):
    end = chr(v2[i] ^ a1[i%8])
    flag +=end

print (flag)

在这里又把a1转换成了assic形式是为了方便异或运算

最后输出了flag

标签：case,break,bmzctf,elrond32,LABEL,a1,a2,result,刷题
来源： https://www.cnblogs.com/murkuo/p/15725804.html